1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating quality factor in a series LCR circuit?

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    My tutor's notes say that quality factor is supposed to be = 2π (Max. energy stored)/(energy dissipated in one cycle) . So for a standard series LCR circuit, he says:
    max energy = 0.5 L I2
    energy lost per cycle = 0.5 R I2 2π/ω .
    ..and at resonance he calculates it to be equal to be [tex] \frac{1}{R} \sqrt{L/C} [\tex]

    2. Relevant equations

    3. The attempt at a solution
    However, this is how I tried to do it:
    max energy = 0.5 C V2
    energy lost per cycle = 0.5 V2 /R x 2π/ω
    ...and at resonance this gives a quality factor of [tex] R \sqrt{C/L} [\tex]

    Erm.... what da heck?
    Last edited: May 29, 2013
  2. jcsd
  3. May 29, 2013 #2
    Latex...y u no work???
  4. May 29, 2013 #3
    I'll post the answers here :
    Q from the 1st bit: 1/R x (L/C)^(1/2)
    Q from my try: R x (C/L)^(1/2)
  5. May 29, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member


    Q = wL/R = 1/wRC, w = 1/sqrt(LC)
    Q = 2pi(peak energy stored within one cycle)/(energy dissipated during that cycle.)

    Now, about your computation: You are using the wrong V_max across the capacitor to compute max stored energy on the capacitor.

    max V across capacitor = V0/wRC if the driving voltage is V0 cos(wt).
    What is dissipation in R in terms of V0?

    Take it from there.
  6. May 30, 2013 #5
    Max Energy: 0.5 C (V/ωRC)2
    Energy lost per cycle: V2/2R x 2π/ω
    ...and this gives the correct value for Q.

    Thank you.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted